A man stands on a weighing machine kept inside a lift. Initially the lift is ascending with the acceleration `'a'` due to which the reading is `W`. Now the lift decends with the same acceleration and reading `10%` of initial. Find the acceleration of lift ?

Options

A. `g/19`
B. `(9g)/11`
C. zero
D. g

Answer

To solve the problem step by step, we need to analyze the situation of a man standing on a weighing machine inside a lift that is accelerating both upwards and downwards.

### Step 1: Understand the Forces Acting on the Man
When the lift is accelerating upwards with acceleration 'a', the effective gravitational force acting on the man is increased. The reading on the weighing machine (normal force, N) is given by:
\[ N = m(g + a) \]
where:
- \( m \) is the mass of the man,
- \( g \) is the acceleration due to gravity.

### Step 2: Write the Equation for the Initial Condition
Given that the reading when the lift is ascending is \( W \):
\[ W = m(g + a) \]
This is our **Equation 1**.

### Step 3: Analyze the Second Condition
When the lift is descending with the same acceleration 'a', the effective gravitational force acting on the man is reduced. The reading on the weighing machine now is:
\[ N' = m(g - a) \]
According to the problem, this reading is 10% of the initial reading:
\[ N' = 0.1W \]
This can be expressed as:
\[ 0.1W = m(g - a) \]
This is our **Equation 2**.

### Step 4: Substitute Equation 1 into Equation 2
From Equation 1, we have:
\[ W = m(g + a) \]
Substituting this into Equation 2 gives:
\[ 0.1(m(g + a)) = m(g - a) \]

### Step 5: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 0.1(g + a) = g - a \]

### Step 6: Rearranging the Equation
Now, let's rearrange this equation:
\[ 0.1g + 0.1a = g - a \]
Bringing all terms involving 'g' to one side and 'a' to the other side, we get:
\[ 0.1g + 0.1a + a = g \]
\[ 0.1g + 1.1a = g \]

### Step 7: Isolate 'g'
Rearranging gives:
\[ g - 0.1g = 1.1a \]
\[ 0.9g = 1.1a \]

### Step 8: Solve for 'a'
Now, we can solve for 'a':
\[ a = \frac{0.9g}{1.1} \]
This simplifies to:
\[ a = \frac{9g}{11} \]

### Final Answer
Thus, the acceleration of the lift is:
\[ a = \frac{9g}{11} \]

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